Problem: Let $R$ be the region in the fourth quadrant that is inside the polar curve $r=2$ and inside the polar curve $r=2+\cos(2\theta)$, as shown in the graph. The curves intersect at $\theta=-\dfrac{\pi}{4}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{4}}^{0}\left(2+\cos(2\theta)\right)^2\,d\theta+\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{4}}4\,d\theta$ (Choice B) B $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{0}\left(\left(2+\cos(2\theta)\right)^2-4\right)d\theta$ (Choice C) C $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{4}}^{0}\left(\left(2+\cos(2\theta)\right)^2-4\right)d\theta$ (Choice D) D $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{-\scriptsize\dfrac{\pi}{4}}\left(2+\cos(2\theta)\right)^2\,d\theta+\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{4}}^{0}4\,d\theta$
Solution: Since we are dealing with two separate polar curves, a good first step is to identify two areas, each enclosed by a single curve, that together define $R$. Such are $R_1$ and $R_2$ : $y$ $x$ $ R_1$ $ R_2$ $ 1$ $ 1$ $R_1$ is enclosed by $r=2+\cos(2\theta)$ and $R_2$ is enclosed by $r=2$. Once we express them as integrals, we can find $R$ using the following relationship: $\text{Area of }R=\text{Area of }R_1+\text{Area of }R_2$ $R_1$ is enclosed by $r=2+\cos(2\theta)$ between $\alpha=-\dfrac{\pi}{2}$ and $\beta=-\dfrac{\pi}{4}$ : $\begin{aligned} \text{Area of }R_1&=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{-\scriptsize\dfrac{\pi}{4}}\left(2+\cos(2\theta)\right)^2\,d\theta \end{aligned}$ $R_2$ is enclosed by $r=2$ between $\alpha=-\dfrac{\pi}{4}$, and $\beta=0$ : $\begin{aligned} \text{Area of }R_2&=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{4}}^{0}\left(2\right)^2\,d\theta \\\\ &=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{4}}^{0}4\,d\theta \end{aligned}$ Now we can express the area of $R$ : $\begin{aligned} &\phantom{=}\text{Area of }R \\\\ &=\text{Area of }R_1+\text{Area of }R_2 \\\\ &=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{-\scriptsize\dfrac{\pi}{4}}\left(2+\cos(2\theta)\right)^2\,d\theta+\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{4}}^{0}4\,d\theta \end{aligned}$